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Murljashka [212]
3 years ago
11

Intr-o clasa snt 35 de elevi .nr fete este egal cu 75 la suta din nr baietilor .aflati nr baietilor

Mathematics
1 answer:
Zanzabum3 years ago
5 0

Salut!

b=baiat

f=fata

In clasa sunt 35 elevi.

75/100 x b=f

3b/4=f

3b=4f

b+f=35

3b+3f=105

7f=105

f=15

b=35-15

b=20

Succes!

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Solve the following by using mathematical Induction. For >/ 1
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Answer:

See explanation

Step-by-step explanation:

Prove that

1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)

1. When n=1, we have

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  • in right part \dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.

2. Assume that for all k following equality is true

1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)

3. Prove that for k+1 the following equality is true too.

1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)

Consider left part:

1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}

Consider right part:

\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)

We get the same left and right parts, so the equality is true for k+1.

By mathematical induction, this equality is true for all n.

3 0
3 years ago
How to find this answer? <br>​
solniwko [45]

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SHEESH this ez 12x20 = 240cm

Step-by-step explanation:

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2 years ago
A four sided sandbox has exactly 2 right angles to side lengths 5 feet what geometric shape as described shape of the sandbox
lana66690 [7]

Answer:

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I have no access at 6 pm
notka56 [123]

Answer:

A and E

Step-by-step explanation:

I would plug in N as any number just to see if they are equal. I did 5.

<u>A</u>. 28(5)-35= -105

<u>B</u>. 28(5-35)= -840

<u>C. </u>28(5-7)= -56

<u>D. </u> 7(4*5-35)= -105

<u>E. </u>7(4*5-5)=  105

4 0
3 years ago
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